llvm-6502/lib/Transforms/Scalar/Reassociate.cpp
2005-08-24 17:55:32 +00:00

665 lines
24 KiB
C++

//===- Reassociate.cpp - Reassociate binary expressions -------------------===//
//
// The LLVM Compiler Infrastructure
//
// This file was developed by the LLVM research group and is distributed under
// the University of Illinois Open Source License. See LICENSE.TXT for details.
//
//===----------------------------------------------------------------------===//
//
// This pass reassociates commutative expressions in an order that is designed
// to promote better constant propagation, GCSE, LICM, PRE...
//
// For example: 4 + (x + 5) -> x + (4 + 5)
//
// In the implementation of this algorithm, constants are assigned rank = 0,
// function arguments are rank = 1, and other values are assigned ranks
// corresponding to the reverse post order traversal of current function
// (starting at 2), which effectively gives values in deep loops higher rank
// than values not in loops.
//
//===----------------------------------------------------------------------===//
#define DEBUG_TYPE "reassociate"
#include "llvm/Transforms/Scalar.h"
#include "llvm/Constants.h"
#include "llvm/Function.h"
#include "llvm/Instructions.h"
#include "llvm/Pass.h"
#include "llvm/Type.h"
#include "llvm/Assembly/Writer.h"
#include "llvm/Support/CFG.h"
#include "llvm/Support/Debug.h"
#include "llvm/ADT/PostOrderIterator.h"
#include "llvm/ADT/Statistic.h"
#include <algorithm>
using namespace llvm;
namespace {
Statistic<> NumLinear ("reassociate","Number of insts linearized");
Statistic<> NumChanged("reassociate","Number of insts reassociated");
Statistic<> NumSwapped("reassociate","Number of insts with operands swapped");
Statistic<> NumAnnihil("reassociate","Number of expr tree annihilated");
struct ValueEntry {
unsigned Rank;
Value *Op;
ValueEntry(unsigned R, Value *O) : Rank(R), Op(O) {}
};
inline bool operator<(const ValueEntry &LHS, const ValueEntry &RHS) {
return LHS.Rank > RHS.Rank; // Sort so that highest rank goes to start.
}
class Reassociate : public FunctionPass {
std::map<BasicBlock*, unsigned> RankMap;
std::map<Value*, unsigned> ValueRankMap;
bool MadeChange;
public:
bool runOnFunction(Function &F);
virtual void getAnalysisUsage(AnalysisUsage &AU) const {
AU.setPreservesCFG();
}
private:
void BuildRankMap(Function &F);
unsigned getRank(Value *V);
void RewriteExprTree(BinaryOperator *I, unsigned Idx,
std::vector<ValueEntry> &Ops);
void OptimizeExpression(unsigned Opcode, std::vector<ValueEntry> &Ops);
void LinearizeExprTree(BinaryOperator *I, std::vector<ValueEntry> &Ops);
void LinearizeExpr(BinaryOperator *I);
void ReassociateBB(BasicBlock *BB);
};
RegisterOpt<Reassociate> X("reassociate", "Reassociate expressions");
}
// Public interface to the Reassociate pass
FunctionPass *llvm::createReassociatePass() { return new Reassociate(); }
static bool isUnmovableInstruction(Instruction *I) {
if (I->getOpcode() == Instruction::PHI ||
I->getOpcode() == Instruction::Alloca ||
I->getOpcode() == Instruction::Load ||
I->getOpcode() == Instruction::Malloc ||
I->getOpcode() == Instruction::Invoke ||
I->getOpcode() == Instruction::Call ||
I->getOpcode() == Instruction::Div ||
I->getOpcode() == Instruction::Rem)
return true;
return false;
}
void Reassociate::BuildRankMap(Function &F) {
unsigned i = 2;
// Assign distinct ranks to function arguments
for (Function::arg_iterator I = F.arg_begin(), E = F.arg_end(); I != E; ++I)
ValueRankMap[I] = ++i;
ReversePostOrderTraversal<Function*> RPOT(&F);
for (ReversePostOrderTraversal<Function*>::rpo_iterator I = RPOT.begin(),
E = RPOT.end(); I != E; ++I) {
BasicBlock *BB = *I;
unsigned BBRank = RankMap[BB] = ++i << 16;
// Walk the basic block, adding precomputed ranks for any instructions that
// we cannot move. This ensures that the ranks for these instructions are
// all different in the block.
for (BasicBlock::iterator I = BB->begin(), E = BB->end(); I != E; ++I)
if (isUnmovableInstruction(I))
ValueRankMap[I] = ++BBRank;
}
}
unsigned Reassociate::getRank(Value *V) {
if (isa<Argument>(V)) return ValueRankMap[V]; // Function argument...
Instruction *I = dyn_cast<Instruction>(V);
if (I == 0) return 0; // Otherwise it's a global or constant, rank 0.
unsigned &CachedRank = ValueRankMap[I];
if (CachedRank) return CachedRank; // Rank already known?
// If this is an expression, return the 1+MAX(rank(LHS), rank(RHS)) so that
// we can reassociate expressions for code motion! Since we do not recurse
// for PHI nodes, we cannot have infinite recursion here, because there
// cannot be loops in the value graph that do not go through PHI nodes.
unsigned Rank = 0, MaxRank = RankMap[I->getParent()];
for (unsigned i = 0, e = I->getNumOperands();
i != e && Rank != MaxRank; ++i)
Rank = std::max(Rank, getRank(I->getOperand(i)));
// If this is a not or neg instruction, do not count it for rank. This
// assures us that X and ~X will have the same rank.
if (!I->getType()->isIntegral() ||
(!BinaryOperator::isNot(I) && !BinaryOperator::isNeg(I)))
++Rank;
//DEBUG(std::cerr << "Calculated Rank[" << V->getName() << "] = "
//<< Rank << "\n");
return CachedRank = Rank;
}
/// isReassociableOp - Return true if V is an instruction of the specified
/// opcode and if it only has one use.
static BinaryOperator *isReassociableOp(Value *V, unsigned Opcode) {
if (V->hasOneUse() && isa<Instruction>(V) &&
cast<Instruction>(V)->getOpcode() == Opcode)
return cast<BinaryOperator>(V);
return 0;
}
/// LowerNegateToMultiply - Replace 0-X with X*-1.
///
static Instruction *LowerNegateToMultiply(Instruction *Neg) {
Constant *Cst;
if (Neg->getType()->isFloatingPoint())
Cst = ConstantFP::get(Neg->getType(), -1);
else
Cst = ConstantInt::getAllOnesValue(Neg->getType());
std::string NegName = Neg->getName(); Neg->setName("");
Instruction *Res = BinaryOperator::createMul(Neg->getOperand(1), Cst, NegName,
Neg);
Neg->replaceAllUsesWith(Res);
Neg->eraseFromParent();
return Res;
}
// Given an expression of the form '(A+B)+(D+C)', turn it into '(((A+B)+C)+D)'.
// Note that if D is also part of the expression tree that we recurse to
// linearize it as well. Besides that case, this does not recurse into A,B, or
// C.
void Reassociate::LinearizeExpr(BinaryOperator *I) {
BinaryOperator *LHS = cast<BinaryOperator>(I->getOperand(0));
BinaryOperator *RHS = cast<BinaryOperator>(I->getOperand(1));
assert(isReassociableOp(LHS, I->getOpcode()) &&
isReassociableOp(RHS, I->getOpcode()) &&
"Not an expression that needs linearization?");
DEBUG(std::cerr << "Linear" << *LHS << *RHS << *I);
// Move the RHS instruction to live immediately before I, avoiding breaking
// dominator properties.
RHS->moveBefore(I);
// Move operands around to do the linearization.
I->setOperand(1, RHS->getOperand(0));
RHS->setOperand(0, LHS);
I->setOperand(0, RHS);
++NumLinear;
MadeChange = true;
DEBUG(std::cerr << "Linearized: " << *I);
// If D is part of this expression tree, tail recurse.
if (isReassociableOp(I->getOperand(1), I->getOpcode()))
LinearizeExpr(I);
}
/// LinearizeExprTree - Given an associative binary expression tree, traverse
/// all of the uses putting it into canonical form. This forces a left-linear
/// form of the the expression (((a+b)+c)+d), and collects information about the
/// rank of the non-tree operands.
///
/// This returns the rank of the RHS operand, which is known to be the highest
/// rank value in the expression tree.
///
void Reassociate::LinearizeExprTree(BinaryOperator *I,
std::vector<ValueEntry> &Ops) {
Value *LHS = I->getOperand(0), *RHS = I->getOperand(1);
unsigned Opcode = I->getOpcode();
// First step, linearize the expression if it is in ((A+B)+(C+D)) form.
BinaryOperator *LHSBO = isReassociableOp(LHS, Opcode);
BinaryOperator *RHSBO = isReassociableOp(RHS, Opcode);
// If this is a multiply expression tree and it contains internal negations,
// transform them into multiplies by -1 so they can be reassociated.
if (I->getOpcode() == Instruction::Mul) {
if (!LHSBO && LHS->hasOneUse() && BinaryOperator::isNeg(LHS)) {
LHS = LowerNegateToMultiply(cast<Instruction>(LHS));
LHSBO = isReassociableOp(LHS, Opcode);
}
if (!RHSBO && RHS->hasOneUse() && BinaryOperator::isNeg(RHS)) {
RHS = LowerNegateToMultiply(cast<Instruction>(RHS));
RHSBO = isReassociableOp(RHS, Opcode);
}
}
if (!LHSBO) {
if (!RHSBO) {
// Neither the LHS or RHS as part of the tree, thus this is a leaf. As
// such, just remember these operands and their rank.
Ops.push_back(ValueEntry(getRank(LHS), LHS));
Ops.push_back(ValueEntry(getRank(RHS), RHS));
return;
} else {
// Turn X+(Y+Z) -> (Y+Z)+X
std::swap(LHSBO, RHSBO);
std::swap(LHS, RHS);
bool Success = !I->swapOperands();
assert(Success && "swapOperands failed");
MadeChange = true;
}
} else if (RHSBO) {
// Turn (A+B)+(C+D) -> (((A+B)+C)+D). This guarantees the the RHS is not
// part of the expression tree.
LinearizeExpr(I);
LHS = LHSBO = cast<BinaryOperator>(I->getOperand(0));
RHS = I->getOperand(1);
RHSBO = 0;
}
// Okay, now we know that the LHS is a nested expression and that the RHS is
// not. Perform reassociation.
assert(!isReassociableOp(RHS, Opcode) && "LinearizeExpr failed!");
// Move LHS right before I to make sure that the tree expression dominates all
// values.
LHSBO->moveBefore(I);
// Linearize the expression tree on the LHS.
LinearizeExprTree(LHSBO, Ops);
// Remember the RHS operand and its rank.
Ops.push_back(ValueEntry(getRank(RHS), RHS));
}
// RewriteExprTree - Now that the operands for this expression tree are
// linearized and optimized, emit them in-order. This function is written to be
// tail recursive.
void Reassociate::RewriteExprTree(BinaryOperator *I, unsigned i,
std::vector<ValueEntry> &Ops) {
if (i+2 == Ops.size()) {
if (I->getOperand(0) != Ops[i].Op ||
I->getOperand(1) != Ops[i+1].Op) {
DEBUG(std::cerr << "RA: " << *I);
I->setOperand(0, Ops[i].Op);
I->setOperand(1, Ops[i+1].Op);
DEBUG(std::cerr << "TO: " << *I);
MadeChange = true;
++NumChanged;
}
return;
}
assert(i+2 < Ops.size() && "Ops index out of range!");
if (I->getOperand(1) != Ops[i].Op) {
DEBUG(std::cerr << "RA: " << *I);
I->setOperand(1, Ops[i].Op);
DEBUG(std::cerr << "TO: " << *I);
MadeChange = true;
++NumChanged;
}
RewriteExprTree(cast<BinaryOperator>(I->getOperand(0)), i+1, Ops);
}
// NegateValue - Insert instructions before the instruction pointed to by BI,
// that computes the negative version of the value specified. The negative
// version of the value is returned, and BI is left pointing at the instruction
// that should be processed next by the reassociation pass.
//
static Value *NegateValue(Value *V, Instruction *BI) {
// We are trying to expose opportunity for reassociation. One of the things
// that we want to do to achieve this is to push a negation as deep into an
// expression chain as possible, to expose the add instructions. In practice,
// this means that we turn this:
// X = -(A+12+C+D) into X = -A + -12 + -C + -D = -12 + -A + -C + -D
// so that later, a: Y = 12+X could get reassociated with the -12 to eliminate
// the constants. We assume that instcombine will clean up the mess later if
// we introduce tons of unnecessary negation instructions...
//
if (Instruction *I = dyn_cast<Instruction>(V))
if (I->getOpcode() == Instruction::Add && I->hasOneUse()) {
Value *RHS = NegateValue(I->getOperand(1), BI);
Value *LHS = NegateValue(I->getOperand(0), BI);
// We must actually insert a new add instruction here, because the neg
// instructions do not dominate the old add instruction in general. By
// adding it now, we are assured that the neg instructions we just
// inserted dominate the instruction we are about to insert after them.
//
return BinaryOperator::create(Instruction::Add, LHS, RHS,
I->getName()+".neg", BI);
}
// Insert a 'neg' instruction that subtracts the value from zero to get the
// negation.
//
return BinaryOperator::createNeg(V, V->getName() + ".neg", BI);
}
/// BreakUpSubtract - If we have (X-Y), and if either X is an add, or if this is
/// only used by an add, transform this into (X+(0-Y)) to promote better
/// reassociation.
static Instruction *BreakUpSubtract(Instruction *Sub) {
// Don't bother to break this up unless either the LHS is an associable add or
// if this is only used by one.
if (!isReassociableOp(Sub->getOperand(0), Instruction::Add) &&
!isReassociableOp(Sub->getOperand(1), Instruction::Add) &&
!(Sub->hasOneUse() &&isReassociableOp(Sub->use_back(), Instruction::Add)))
return 0;
// Convert a subtract into an add and a neg instruction... so that sub
// instructions can be commuted with other add instructions...
//
// Calculate the negative value of Operand 1 of the sub instruction...
// and set it as the RHS of the add instruction we just made...
//
std::string Name = Sub->getName();
Sub->setName("");
Value *NegVal = NegateValue(Sub->getOperand(1), Sub);
Instruction *New =
BinaryOperator::createAdd(Sub->getOperand(0), NegVal, Name, Sub);
// Everyone now refers to the add instruction.
Sub->replaceAllUsesWith(New);
Sub->eraseFromParent();
DEBUG(std::cerr << "Negated: " << *New);
return New;
}
/// ConvertShiftToMul - If this is a shift of a reassociable multiply or is used
/// by one, change this into a multiply by a constant to assist with further
/// reassociation.
static Instruction *ConvertShiftToMul(Instruction *Shl) {
if (!isReassociableOp(Shl->getOperand(0), Instruction::Mul) &&
!(Shl->hasOneUse() && isReassociableOp(Shl->use_back(),Instruction::Mul)))
return 0;
Constant *MulCst = ConstantInt::get(Shl->getType(), 1);
MulCst = ConstantExpr::getShl(MulCst, cast<Constant>(Shl->getOperand(1)));
std::string Name = Shl->getName(); Shl->setName("");
Instruction *Mul = BinaryOperator::createMul(Shl->getOperand(0), MulCst,
Name, Shl);
Shl->replaceAllUsesWith(Mul);
Shl->eraseFromParent();
return Mul;
}
// Scan backwards and forwards among values with the same rank as element i to
// see if X exists. If X does not exist, return i.
static unsigned FindInOperandList(std::vector<ValueEntry> &Ops, unsigned i,
Value *X) {
unsigned XRank = Ops[i].Rank;
unsigned e = Ops.size();
for (unsigned j = i+1; j != e && Ops[j].Rank == XRank; ++j)
if (Ops[j].Op == X)
return j;
// Scan backwards
for (unsigned j = i-1; j != ~0U && Ops[j].Rank == XRank; --j)
if (Ops[j].Op == X)
return j;
return i;
}
void Reassociate::OptimizeExpression(unsigned Opcode,
std::vector<ValueEntry> &Ops) {
// Now that we have the linearized expression tree, try to optimize it.
// Start by folding any constants that we found.
bool IterateOptimization = false;
if (Ops.size() == 1) return;
if (Constant *V1 = dyn_cast<Constant>(Ops[Ops.size()-2].Op))
if (Constant *V2 = dyn_cast<Constant>(Ops.back().Op)) {
Ops.pop_back();
Ops.back().Op = ConstantExpr::get(Opcode, V1, V2);
OptimizeExpression(Opcode, Ops);
return;
}
// Check for destructive annihilation due to a constant being used.
if (ConstantIntegral *CstVal = dyn_cast<ConstantIntegral>(Ops.back().Op))
switch (Opcode) {
default: break;
case Instruction::And:
if (CstVal->isNullValue()) { // ... & 0 -> 0
Ops[0].Op = CstVal;
Ops.erase(Ops.begin()+1, Ops.end());
++NumAnnihil;
return;
} else if (CstVal->isAllOnesValue()) { // ... & -1 -> ...
Ops.pop_back();
}
break;
case Instruction::Mul:
if (CstVal->isNullValue()) { // ... * 0 -> 0
Ops[0].Op = CstVal;
Ops.erase(Ops.begin()+1, Ops.end());
++NumAnnihil;
return;
} else if (cast<ConstantInt>(CstVal)->getRawValue() == 1) {
Ops.pop_back(); // ... * 1 -> ...
}
break;
case Instruction::Or:
if (CstVal->isAllOnesValue()) { // ... | -1 -> -1
Ops[0].Op = CstVal;
Ops.erase(Ops.begin()+1, Ops.end());
++NumAnnihil;
return;
}
// FALLTHROUGH!
case Instruction::Add:
case Instruction::Xor:
if (CstVal->isNullValue()) // ... [|^+] 0 -> ...
Ops.pop_back();
break;
}
// Handle destructive annihilation do to identities between elements in the
// argument list here.
switch (Opcode) {
default: break;
case Instruction::And:
case Instruction::Or:
case Instruction::Xor:
// Scan the operand lists looking for X and ~X pairs, along with X,X pairs.
// If we find any, we can simplify the expression. X&~X == 0, X|~X == -1.
for (unsigned i = 0, e = Ops.size(); i != e; ++i) {
// First, check for X and ~X in the operand list.
if (BinaryOperator::isNot(Ops[i].Op)) { // Cannot occur for ^.
Value *X = BinaryOperator::getNotArgument(Ops[i].Op);
unsigned FoundX = FindInOperandList(Ops, i, X);
if (FoundX != i) {
if (Opcode == Instruction::And) { // ...&X&~X = 0
Ops[0].Op = Constant::getNullValue(X->getType());
Ops.erase(Ops.begin()+1, Ops.end());
++NumAnnihil;
return;
} else if (Opcode == Instruction::Or) { // ...|X|~X = -1
Ops[0].Op = ConstantIntegral::getAllOnesValue(X->getType());
Ops.erase(Ops.begin()+1, Ops.end());
++NumAnnihil;
return;
}
}
}
// Next, check for duplicate pairs of values, which we assume are next to
// each other, due to our sorting criteria.
if (i+1 != Ops.size() && Ops[i+1].Op == Ops[i].Op) {
if (Opcode == Instruction::And || Opcode == Instruction::Or) {
// Drop duplicate values.
Ops.erase(Ops.begin()+i);
--i; --e;
IterateOptimization = true;
++NumAnnihil;
} else {
assert(Opcode == Instruction::Xor);
if (e == 2) {
Ops[0].Op = Constant::getNullValue(Ops[0].Op->getType());
Ops.erase(Ops.begin()+1, Ops.end());
++NumAnnihil;
return;
}
// ... X^X -> ...
Ops.erase(Ops.begin()+i, Ops.begin()+i+2);
i -= 1; e -= 2;
IterateOptimization = true;
++NumAnnihil;
}
}
}
break;
case Instruction::Add:
// Scan the operand lists looking for X and -X pairs. If we find any, we
// can simplify the expression. X+-X == 0
for (unsigned i = 0, e = Ops.size(); i != e; ++i) {
// Check for X and -X in the operand list.
if (BinaryOperator::isNeg(Ops[i].Op)) {
Value *X = BinaryOperator::getNegArgument(Ops[i].Op);
unsigned FoundX = FindInOperandList(Ops, i, X);
if (FoundX != i) {
// Remove X and -X from the operand list.
if (Ops.size() == 2) {
Ops[0].Op = Constant::getNullValue(X->getType());
Ops.erase(Ops.begin()+1);
++NumAnnihil;
return;
} else {
Ops.erase(Ops.begin()+i);
if (i < FoundX) --FoundX;
Ops.erase(Ops.begin()+FoundX);
IterateOptimization = true;
++NumAnnihil;
}
}
}
}
break;
//case Instruction::Mul:
}
if (IterateOptimization)
OptimizeExpression(Opcode, Ops);
}
/// PrintOps - Print out the expression identified in the Ops list.
///
static void PrintOps(unsigned Opcode, const std::vector<ValueEntry> &Ops,
BasicBlock *BB) {
Module *M = BB->getParent()->getParent();
std::cerr << Instruction::getOpcodeName(Opcode) << " "
<< *Ops[0].Op->getType();
for (unsigned i = 0, e = Ops.size(); i != e; ++i)
WriteAsOperand(std::cerr << " ", Ops[i].Op, false, true, M)
<< "," << Ops[i].Rank;
}
/// ReassociateBB - Inspect all of the instructions in this basic block,
/// reassociating them as we go.
void Reassociate::ReassociateBB(BasicBlock *BB) {
for (BasicBlock::iterator BI = BB->begin(); BI != BB->end(); ++BI) {
if (BI->getOpcode() == Instruction::Shl &&
isa<ConstantInt>(BI->getOperand(1)))
if (Instruction *NI = ConvertShiftToMul(BI)) {
MadeChange = true;
BI = NI;
}
// Reject cases where it is pointless to do this.
if (!isa<BinaryOperator>(BI) || BI->getType()->isFloatingPoint())
continue; // Floating point ops are not associative.
// If this is a subtract instruction which is not already in negate form,
// see if we can convert it to X+-Y.
if (BI->getOpcode() == Instruction::Sub) {
if (!BinaryOperator::isNeg(BI)) {
if (Instruction *NI = BreakUpSubtract(BI)) {
MadeChange = true;
BI = NI;
}
} else {
// Otherwise, this is a negation. See if the operand is a multiply tree
// and if this is not an inner node of a multiply tree.
if (isReassociableOp(BI->getOperand(1), Instruction::Mul) &&
(!BI->hasOneUse() ||
!isReassociableOp(BI->use_back(), Instruction::Mul))) {
BI = LowerNegateToMultiply(BI);
MadeChange = true;
}
}
}
// If this instruction is a commutative binary operator, process it.
if (!BI->isAssociative()) continue;
BinaryOperator *I = cast<BinaryOperator>(BI);
// If this is an interior node of a reassociable tree, ignore it until we
// get to the root of the tree, to avoid N^2 analysis.
if (I->hasOneUse() && isReassociableOp(I->use_back(), I->getOpcode()))
continue;
// First, walk the expression tree, linearizing the tree, collecting
std::vector<ValueEntry> Ops;
LinearizeExprTree(I, Ops);
DEBUG(std::cerr << "RAIn:\t"; PrintOps(I->getOpcode(), Ops, BB);
std::cerr << "\n");
// Now that we have linearized the tree to a list and have gathered all of
// the operands and their ranks, sort the operands by their rank. Use a
// stable_sort so that values with equal ranks will have their relative
// positions maintained (and so the compiler is deterministic). Note that
// this sorts so that the highest ranking values end up at the beginning of
// the vector.
std::stable_sort(Ops.begin(), Ops.end());
// OptimizeExpression - Now that we have the expression tree in a convenient
// sorted form, optimize it globally if possible.
OptimizeExpression(I->getOpcode(), Ops);
// We want to sink immediates as deeply as possible except in the case where
// this is a multiply tree used only by an add, and the immediate is a -1.
// In this case we reassociate to put the negation on the outside so that we
// can fold the negation into the add: (-X)*Y + Z -> Z-X*Y
if (I->getOpcode() == Instruction::Mul && I->hasOneUse() &&
cast<Instruction>(I->use_back())->getOpcode() == Instruction::Add &&
isa<ConstantInt>(Ops.back().Op) &&
cast<ConstantInt>(Ops.back().Op)->isAllOnesValue()) {
Ops.insert(Ops.begin(), Ops.back());
Ops.pop_back();
}
DEBUG(std::cerr << "RAOut:\t"; PrintOps(I->getOpcode(), Ops, BB);
std::cerr << "\n");
if (Ops.size() == 1) {
// This expression tree simplified to something that isn't a tree,
// eliminate it.
I->replaceAllUsesWith(Ops[0].Op);
} else {
// Now that we ordered and optimized the expressions, splat them back into
// the expression tree, removing any unneeded nodes.
RewriteExprTree(I, 0, Ops);
}
}
}
bool Reassociate::runOnFunction(Function &F) {
// Recalculate the rank map for F
BuildRankMap(F);
MadeChange = false;
for (Function::iterator FI = F.begin(), FE = F.end(); FI != FE; ++FI)
ReassociateBB(FI);
// We are done with the rank map...
RankMap.clear();
ValueRankMap.clear();
return MadeChange;
}